MI 2022 PU1 H2 CHEM EOY

Class Adm No Candidate Name: This question paper consists of 17 printed pages and 1 blank page. 2022 End-of-Year Examination Pre-University 1 H2 CHEMISTRY 9729/01 Paper 1 Multiple Choice & Structured Questions 12 Oct 2022 2 hours Additional materials: Multiple Choice Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST Do not turn over this question paper until you are told to do so Write your name, class and admission number on all the work you hand in. For Section A, there are fifteen questions. Answer ALL questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the Multiple Choice Answer Sheet provided. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Read the instructions on the Multiple Choice Answer Sheet very carefully. For Section B, write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all the questions. The use of an approved scientific calculator is expected , where appropriate. Any rough working should be done in this question paper. Question Section A Section B Total 1 2 3 4 5 6 Marks 15 10 12 10 8 9 6 70

2 DA 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 A 8.3% 5.8% 5.8% 28.9% 71.1% 16.5% 14.0% 9.9% 0.0% 9.9% 18.2% 8.3% 24.0% 15.7% 24.0% B 6.6% 72.7% 19.8% 7.4% 15.7% 4.1% 51.2% 14.9% 0.0% 32.2% 40.5% 18.2% 63.6% 43.0% 36.4% C 36.4% 12.4% 21.5% 44.6% 8.3% 69.4% 23.1% 11.6% 57.0% 14.0% 26.4% 62.0% 10.7% 9.1% 25.6% D 48.8% 9.1% 52.1% 19.0% 4.1% 9.9% 9.9% 63.6% 43.0% 40.5% 13.2% 10.7% 1.7% 29.8% 13.2% Qn Distractor 4 C – Higher Mr of ethanoic acid is due to formation of dimer via H-bonding in non-polar benzene solvent. 7 B – Did not account for original 8 dm3 in new total volume C – Did not account for the 9 dm3 added to new total volume 9 C – Careless, 1 mol must compare to 1 mol

3 [Turn over Section A – Multiple Choice For each question there are four possible answers, A, B, C, and D. Choose the one you consider to be correct. 1 How many subshells and orbitals are there in principal quantum shell number 3? subshells orbitals A 2 4 B 2 6 C 3 6 D 3 9 3s 3p 3d ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂ 2 Use of the Data Booklet is relevant to this question. The following are flight paths of charged particles when accelerated in an electric field. Which correctly identifies S, T and U? S T U A 14N+ 14C– 14C2– B 14N– 12C+ 12C2+ C 12C– 14N2+ 14N+ D 14C– 12C+ 14N+ S must be negatively charged (attracted to positive terminal), T & U must be positively charged (attracted to negative terminal). U must have a larger 𝑞 𝑚 ratio compared to T due to its larger deflection angle.

4 3 In microwave ovens, the energy produced is absorbed by polar molecules. Which of the following would absorb microwave energy? 1 CO2 2 BrF3 3 BF3 4 SO2 A 1 and 2 only B 1 and 3 only C 2 and 3 only D 2 and 4 only To determine if a molecule is polar, the following must be done in sequence: 1. Draw structure with correct shape (if not familiar, you may have to start from dot-and-cross) 2. Identify all individual bond dipoles 3. Check for net dipole moment CO2 BrF3 BF3 SO2 2bp 0lp 3bp 2lp 3bp 0lp 2bp 1lp linear T-shape trigonal planar bent

5 [Turn over 4 Which of the following cannot be explained by hydrogen bonding? A the difference in boiling point between ethanol and hexan-1-ol B the difference in melting point between H2O and HF C the higher than expected relative molecular mass of ethanoic acid in benzene D the difference in density between water and ice Boiling point: ethanol < hexan-1-ol Same average no. of H -bonds per molecule (1), also same polarity of O –H bond. Similar H- bonding strength. Difference in boiling point is due to large electron cloud size of hexan -1-ol leading to stronger id-id IMFOA. Melting point: H2O > HF H2O has an average of 2 H -bonds per molecule compared to the 1 of HF, hence stronger H - bonding. Higher Mr of ethanoic acid is due to formation of dimer via H -bonding in non -polar benzene solvent. Density: water > ice Lower density of ice is due to its open structure, which results from the orderly tetrahedral arrangement of its molecules H-bonded to each other. 5 Trifluorooxonium has the formula OF3n+ and its shape is trigonal pyramidal. What is the value of n in trifluorooxonium? A 1 B 2 C 3 D 4 F atoms will only form single bonds to O. Trigonal pyramidal → 3bp 1lp → 5 val e– on O → 1 less than usual 6, thus n=1.

6 6 Histamine is produced in the body to help fight infection. Its shape allows it to fit into receptors which expand blood vessels. histamine What are the values of the bond angles x, y and z? x y z A 120° 90° 120° B 109° 109° 107° C 107° 109° 120° D 107° 120° 109° x – 3bp 1lp (trigonal pyramidal) y – 4bp 0lp (tetrahedral) z – 3bp 0lp (trigonal planar) 7 Two bulbs R and S, containing Ne and Ar respectively, are connected to a 9 dm 3 vacuum chamber as shown. What will be the total pressure in the vessel when the valves are opened at constant temperature? A 168 kPa B 317 kPa C 356 kPa D 375 kPa nT = nR + nS at constant T, PTVT = PRVR + PSVS PT(17.0 dm3) = (300 kPa)(5.0 dm3) + (450 kPa)(3.0 dm3) PT = 168 kPa (3sf)

7 [Turn over 8 In which reactions does NH3 behave as a Brønsted-Lowry acid? 1 HSO4– + NH3 → SO42– + NH4+ 2 Ag+ + 2NH3 → [Ag(NH3)2]+ 3 NH3 + PO43– → NH2– + HPO42– A 1 and 2 only B 1 and 3 only C 2 and 3 only D 3 only 1: NH3 becomes NH4+ (H+ acceptor, base) 2: NH3 stays as NH3 (not acid-base reaction) 3: NH3 becomes NH2– (H+ donor, acid) 9 Which of the following correctly defines the term relative atomic mass of an element? A the mass of 1 atom of an element relative to the mass of 1 atom of 12C B the mass of 1 mole of atoms of an element divided by 6.02 x 1023 C the mass of 1 mole of atoms of an element relative to 1 12 the mass of 1 atom of 12C D the mass of 1 mole of atoms of an element relative to 1 12 the mass of 1 mole of 12C atoms Definition question. 10 A carbon sample contains a mixture of 12C and 14C isotopes. When 1.000 g of this sample is burned completely in 16O2, the mass of CO2 formed is 3.55 g. What is the percentage by mass of the 12C isotope in this sample? A 12.4% B 30.6% C 50.5% D 69.4% 12C + O2 → 12CO2 14C + O2 → 14CO2 Mass of C / g 𝑥 1-𝑥 Amount of C / mol 𝑥 12 1−𝑥 14 Amount of CO2 / mol 𝑥 12 1−𝑥 14 Mass of CO2 formed = Mass of 12CO2 + Mass of 14CO2 3.55 = [ 𝑥 12 × (12 + 32)] + [ 1−𝑥 14 × (14 + 32)] 3.55 = 44 12 𝑥 + 46 14 − 46 14 𝑥 𝑥 = 0.6937 g % by mass of 12C = 0.6937 1 × 100% = 69.4% (3sf)

8 11 An ion of metal M can be oxidised by potassium manganate(VII) in acid solution to form MO3−. In an experiment, 0.00500 mol of the ion of M required 15.0 cm3 of 0.200 mol dm−3 potassium manganate(VII) for complete reaction. What is the initial oxidation state of the ion of M given that potassium manganate(VII) is reduced to Mn2+? A +1 B +2 C +4 D +7 [R]: MnO4– + 8H+ + 5e– → Mn2+ + 4H2O [O]: Mn